Since \(r=\lambda (k+1)\)  and \(\left|G_{x}\right|=r\left|G_{x,B}\right|\) , it follows that
\(\left|X_{x}\right|=\frac{\lambda (k+1)\left|G_{x,B}\right|}{\left[ G_{x}:X_{x}\right] } \textit {.} \) 
Assume that \(\left( \Phi _{ef}^{\ast }(p),\left|X_{x}\right|\right) >1\) . If \(\left( \Phi _{ef}^{\ast }(p),k+1\right) =1\) , then \(\left(\Phi _{ef}^{\ast }(p),\lambda \left|G_{x,B}\right|\right) >1\)  and
hence \(\lambda \left|G_{x,B}\right|\geqslant \left[ G_{x}:H_{x}\right] \) 
by Lemma REF . Thus \(\left|X_{x}\right|\geqslant k+1\)  by (REF ), and hence \(X_{x}\)  is a large subgroup of \(X\) , as \(k^{2}=[X:X_{x}]\) .
Since \(\lambda \left|G_{x,B}\right|\mid \left|G_{B}\right|\)  and \(\left( \Phi _{ef}^{\ast }(p),\lambda \left|G_{x,B}\right|\right) >1\) , it follows that \(\left( \Phi _{ef}^{\ast }(p),\left|G_{B}\right|\right) >1\) . Moreover, as \(\Phi _{ef}^{\ast }(p)\equiv 1 \pmod {ef}\)  by [1]}, Proposition 5.2.15.(i), it follows
that \(\left( \Phi _{ef}^{\ast }(p),ef\right) =1\)  and hence \(\left( \Phi _{ef}^{\ast }(p),\left|\mathrm {Out(}X\mathrm {)}\right|\right) =1\) 
(see [1]}, Table 5.1.A). Thus \(\left( \Phi _{ef}^{\ast }(p),\left|X_{B}\right|\right) >1\) , as \(\left[ G_{B}:X_{B}\right]\mid \left|\mathrm {Out(}X\mathrm {)}\right|\) , which is the
assertion (1a).
Assume that \(\left( \Phi _{ef}^{\ast }(p),\left|X_{x}\right|\right) =1\) . Then \(\Phi _{ef}^{\ast }(p)\mid k^{2}\) , since \(\left[ X:X_{x}\right] =k^{2}\) , and hence \(\Phi _{ef}^{\ast }(p)\mid \left|X\right|\) . Arguing as above, we obtain \(\left( \Phi _{ef}^{\ast }(p),\left|X_{B}\right|\right) =\Phi _{ef}^{\ast }(p)^{1/2}\) , which is the assertion (2).
